From the given data: \(\overline{x}_{1}=\frac{\sum x_{1}}{n_{1}}=\frac{78.95}{12}=6.579\)

\(s_{1} = 0.4992\)

\(\overline{x}_{2} = \overline{x}_{1}=\frac{\sum x_{2}}{n_{2}}=\frac{71.47}{11}=6.497\)

\(s_{2} = 0.4791\) Calculating pooled standard deviation: \(s_{p}=\sqrt{\frac{s_{1}^{2}(n_{1}-1)+s_{2}^{2}(n_{2}-1)}{n_{1}+n_{2}-2}} = \sqrt{\frac{0.4992^{2}\times 11+0.4791^{2}\times 10}{12+11-2}=0.4897}\)

\(H0: \mu_{1} = \mu_{2}\)

\(HA: \mu_{1} \neq \mu_{2}\)

a) Leve of significance=0.05

b) Test Statistics: \(t=\frac{\overline{x}_{1}-\overline{x}_{2}}{s_{p}\times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}=\frac{6.579-6.497}{0.4897\times\sqrt{\frac{1}{12}+\frac{1}{11}}}=0.40\)

c)P-value for a two tailed test: Degree of freedom \(= n_{1} + n_{2} - 2 = 12 + 11 - 2 = 21\) Using t-distribution table P-value \(= 0.6932\) As P-value is more than the significance level, we do not have sufficient evidence to reject H0 Conclusion: There is no difference between the yield distributions.